# Square Roots By Long Division

So we can perform long division, and that can be done algebraically to let us perform division of polynomials, computing a quotient and remainder.

So now on to square roots.

We do the same thing. We remove as much as we can from the leading portion, hang on to the remainder, and then carry on. At each stage we have an equation giving the current status.

Let's compute the square root of 177241.

We start by thinking of it as $17*10^4+7241.$ That means the square root will be 400 and a bit (because the square root of 17 is 4 and a bit, and the square root of $10^4$ is 100.)

So $\sqrt{177241}=500$ and a bit.

Let's put that in a table:


4    .    .
+---------------
\| 1 7  7 2  4 1
1 6
-----
1  7 2  4 1

So, specifically, $177241=400^2+17241.$

Now let's go for the next phase. We compute the square root of 1772 as 40 and a bit. So we get $(40+x)^2=1600+2\times40\times{x}+x^2.$ We've already removed the 1600 to be left with 172. That has to be $(2\times40+x)\times{x}.$ Our 40 is the digit at the top of the table, multiply by 2, then put the "x" on the end to make "8x".


4    x    .
+---------------
\| 1 7  7 2  4 1
1 6
-----
8 x     1  7 2

So we want an "x" such that "8x" times "x" is as close as possible to, but no bigger than, 172. Clearly the best value is "2". We insert the value "2", multiply, and subtract:


4    2    .
+---------------
\| 1 7  7 2  4 1
1 6
-----
8 2     1  7 2
1  6 4
--------
8  4 1

Now again, we double the answer so far (giving 84) and find a digit that we can put on the end, then multiply by, to get something as close as possible but no bigger than 841.

Again, the answer is obvious, and in this case is simply 1.


4    2    1
+---------------
\| 1 7  7 2  4 1
1 6
-----
8 2     1  7 2
1  6 4
--------
8 4 1        8  4 1
8  4 1
--------
0

Remainder is zero.

Let's do that with 2:


1 .   4    1    4    2
+-------------------------------
\| 2 . 0 0  0 0  0 0  0 0  0 0
1
---
1 . 0 0
2 4      . 9 6
---------
4  0 0
2 8 1        2  8 1
--------
1  1 9  0 0
2 8 2 4      1  1 2  9 6
-------------
6  0 4  0 0
2 8 2 8 2         5  6 5  6 4
-------------
3 8  3 6  0 0

Here we are looking for the next digit. Our equation so far is that $\sqrt{2}=(1.4142+x)^2.$ We are part way, and we can see that $2=(1.4142)^2+0.00003836.$

We double the bit we have so far, multiply by 10, then find the digit to put on the end, multiply by, and get close to 0.00383600.

So "28284x" times "x" has to be close to, but not exceed, 383600. The value for "x" has to be 1, and so we continue.

And so, back to the original question.

• Is the product of 4 consecutive
positive integers always one
less than a square?
Let's take the product of 4 consecutive positive integers:

• $(n-1)n(n+1)(n+2)=n^4+2n^3-n^2-2n$

• $(n-1)n(n+1)(n+2)+1=n^4+2n^3-n^2-2n+1$
And now we take the square root:


n^2
+------------------------------
\| n^4 + 2n^3 - n^2 - 2n + 1
n^2       n^4
-----
0 + 2n^3 - n^2

Remember, we double what we have so far, and then work out what the next term will be:


n^2 +  X
+------------------------------
\| n^4 + 2n^3 - n^2 - 2n + 1
n^2           n^4
-----
2n^2 + X        0 + 2n^3 - n^2

Whatever X is going to be, we have to get rid of the $2n^3$ term when we multiply by $2n^2,$ so there's only one option - X has to be n.


n^2 +  n
+------------------------------
\| n^4 + 2n^3 - n^2 - 2n + 1
n^2           n^4
-----
2n^2 + n        0 + 2n^3 - n^2
2n^3 + n^2
------------
-2n^2 - 2n + 1

Double the answer so far, and work out what to extend by:


n^2 +  n + X
+------------------------------
\| n^4 + 2n^3 - n^2 - 2n + 1
n^2           n^4
-----
2n^2 + n        0 + 2n^3 - n^2
2n^3 + n^2
------------
2n^2 + 2n + X            -2n^2 - 2n + 1

We can see that X has to be -1 and we get this:


n^2 +  n - 1
+------------------------------
\| n^4 + 2n^3 - n^2 - 2n + 1
n^2           n^4
-----
2n^2 + n        0 + 2n^3 - n^2
2n^3 + n^2
------------
2n^2 + 2n - 1            -2n^2 - 2n + 1
-2n^2 - 2n + 1
----------------
0


... and there's no remainder. Therefore

• $\sqrt{n^4+2n^3-n^2-2n+1}=n^2+n-1$
And we're done.

So, a final example. Let's take the square root of this:


,---------------------------------
v  4n^4 - 12n^3 - 11n^2 + 30n + 24




2n^2 -  3n - 5
,---------------------------------
v  4n^4 - 12n^3 - 11n^2 + 30n + 24
4n^4
------
0 - 12n^3 - 11n^2
4n^2 - 3n          - 12n^3 +  9n^2
-------------------
- 20n^2 + 30n + 24
4n^2 - 6n - 5              - 20n^2 + 30n + 25
--------------------
-  1



So $4n^4-12n^3-11n^2+30n+24=(2n^2-3n-5)^2-1$ and hence it's the difference of two squares. That means we have the factorisation:

$4n^4-12n^3-11n^2+30n+24=(2n^2-3n-5+1)\times(2n^2-3n-5-1)$

or

$4n^4-12n^3-11n^2+30n+24=(2n^2-3n-4)\times(2n^2-3n-6)$